What is the integral from-pi/4 to 0 of sec(x)tan(x) ?

The integral from-pi/4 to 0 of sec(x)tan(x) is -sqrt(2)+1

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integral from-pi/4 to 0 of sec(x)tan(x)

\int_{- \frac{π}{4}}^{0} sec (x) tan (x) d x

- 2 + 1

Decimal

- 0.41421 \dots

Solution steps

\int_{- \frac{π}{4}}^{0} sec (x) tan (x) d x

Apply u-substitution

: \int_{2}^{1} 1 d u

= \int_{2}^{1} 1 d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} 1 d u

Integral of a constant:

\int a d x = a x

= - [1 \cdot u]_{1}^{2}

Simplify

= - [u]_{1}^{2}

Compute the boundaries:

2 - 1

= - (2 - 1)

Simplify

= - 2 + 1

\int_{1}^{5} 7 - 2 x d x

\int_{1}^{4} (1 + \frac{1}{x} + \frac{1}{x ^{2}}) d x

\int_{0}^{1} 3 x + 5 d x

\int_{0}^{t} (t - w) e^{5 w} d w

\int_{0}^{4} e^{x} x^{2} d x

What is the integral from-pi/4 to 0 of sec(x)tan(x) ?
The integral from-pi/4 to 0 of sec(x)tan(x) is -sqrt(2)+1